First, draw the x-y plane. Then, draw the vector <-5,-8>, and it should be in the third quadrant. What the question is asking for is the angle between the positive x axis and that vector, starting at the positive x axis and going counterclockwise. You know from the drawing that tan(θ) = (-8)/(-5), so θ = arctan(8/5), which is about 58 degrees. However, if you look at the picture, you should see that θ is bigger than 180 degrees because it's in the third quadrant.
arctan doesn't tell the whole story here because multiple angles can have the value for tangent. and that's why it's important to draw a picture. The way to interpret the value of 58 degrees for arctan is that if you look at the vector <-5, -8>, it is 58 degrees further counterclockwise than the 180 degree mark, which puts the angle for the vector <-5, -8> at around 238 degrees.